Re: because g⤨(g⁻¹(x)) = g(y) [1/2] Re: how

On 5/10/2024 11:32 PM, Moebius wrote:

Am 11.05.2024 um 01:44 schrieb Jim Burns:

For ω
I suggest the definition:
γ < ω  ⟺
∀β: 0 < β ≤ γ  ⇒
∃α: 0 ≤ α < γ  ∧  α+1 = β >
I'm pretty sure we can derive induction from that.

>
Incidentally, I'm just studying some (early) papers
concerning this question.
>
E. Zermelo, Sur les ensembles finis et
le principe de l'induction complète (1909)
>
K Grelling, Die Axiome der Arithmetik mit
besonderer Berücksichtigung der
Beziehungen zur Mengenlehre (1910)
>
K Grelling, Mengenlehre (1924)
>
But all these considerations presuppose
some sort of set theory.

I'm not shocked.
I think our happy, little band here is
an outlier, when it comes to ideas about
how controversial set theory is.
I also would not be shocked if
they had presupposed arithmetic closure
(something which WM disputes).
----
Suppose, though, that
we don't have sets, but
we have ordinals.
Let's prove cisfinite induction.
Ordinals are well.ordered.
∃γ: P(γ) ⟹ ∃β: (P(β) ∧ ¬∃α<β:P(α))
Define the _finite_ ordinals as
those which can be stepped.back.from and also
each ordinal.before can be stepped.back.from,
except for 0
I'm abbreviating finiteness as '< ω'
γ < ω  ⟺
∀β: 0<β≤γ ⟹ ∃α: (0≤α<β ∧ α+1=β)
| Assume γ < ω exists such that
| P(0) ∧ ¬P(γ)
|
| There is a first β ≤ γ such that ¬P(β)
| β ≠ 0
|
| β ≤ γ is finite.
| β can be stepped.back.from to α
| α+1 = β
| β is the first such that ¬P(β)
| ¬(¬P(α))
|
| Exists α < ω such that P(α) ∧ ¬P(α+1)
Therefore,
if exists γ < ω : P(0) ∧ ¬P(γ)
then exists α < ω : P(α) ∧ ¬P(α+1)
That can be re.written as
P(0) ∧ ∀α<ω: P(α)⇒P(α+1)  ⟹  ∀γ<ω: P(γ)
That is cisfinite induction.